Australian Open 2020: Alexander Zverev vs Stan Wawrinka, quarterfinal match preview and prediction
Alexander Zverev has had more setbacks in hardcourt Slams than anyone would have deserved at the age of just 22. But putting all of that aside, he has finally made quarterfinals of a Slam played on hardcourts, in his 10th try.
And he has done that with some level of conviction. While he might not have come across a big name in the draw just yet, Zverev has still shown real character with wins against the likes of Andrey Rublev and Fernando Verdasco.
The German No. 1's biggest challenge heading into the tournament was to regain the self-belief that had taken a hit after a string of shock losses on the big stage. Having done just that, his reward is a last eight match with former champion Stan Wawrinka.
Now Wawrinka, unlike Zverev, has had a few front-page worthy wins over top players in the draw - the most recent of which was an absolute showcase against Daniil Medvedev.
Wawrinka wasn't in the best form coming into the tournament, but gave himself enough time on the court to get to a point where he is now looking strong in all departments.
He has had good practice at facing big serves coming from Medvedev and to some extent John Isner in the third round, and has delivered solid results in terms of percentage of points won on return (35-45 in both matches).
Now that's an important stat considering how the Zverev serve has turned into his biggest weapon this week. The German has delivered behind the serve and he peaked in the fourth round match against Rublev, winning a mind-boggling 91% of the points behind the first serve.
Despite Zverev's very potent backhand, Wawrinka will have a definitive edge if the rallies. If he can manage to keep the ball in play off of Zverev's serve and put some pressure from the baseline, I could see the German's game breaking down quickly.
While Zverev's run has been inspired so far, he might find the Wawrinka wall insurmountable at the end of this one.
Prediction: Wawrinka to win in 4 sets.